1. ## Limit

Consider a function $\displaystyle f$ that $\displaystyle |f(x) - Ax| \leq e^{-x}$, for any number A $\displaystyle \in$ R. Calculate:

$\displaystyle lim_x{ \to +\infty} \frac{f(x) - f(1)}{x-1}$

2. Hi

$\displaystyle \frac{|f(x) - f(1)|}{|x-1|} = \frac{|(f(x) - Ax) + (Ax - A) + (A - f(1))|}{|x-1|}$

$\displaystyle \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax| + |Ax - A| + |A - f(1)|}{|x-1|}$

Spoiler:

$\displaystyle \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax|}{|x-1|} + \frac{|Ax - A|}{|x-1|} + \frac{|A - f(1)|}{|x-1|}$

$\displaystyle \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}$

And

$\displaystyle \lim_{x \rightarrow +\infty} \left(\frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}\right) = |A|$

$\displaystyle \forall A \in \mathbb{R} ~ \lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} \leq |A|$

Therefore $\displaystyle \lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} = 0$

It would be better if someone could confirm !

3. Thank you men